$\begin{cases}c(1)=56\\\\ c(n)=c(n-1)\cdot \dfrac{1}{2} \end{cases}$ What is the $4^{\text{th}}$ term in the sequence?
Answer: This is a recursive formula. It tells us that the first term is $56$ and that the common ratio is $\dfrac{1}{2}$. $\begin{aligned} {c(1)}&=56 \\\\ {c(2)}&={c(1)}\cdot \dfrac{1}{2}=28 \\\\ {c(3)}&={c(2)}\cdot \dfrac{1}{2}=14 \\\\ {c(4)}&={c(3)}\cdot \dfrac{1}{2}=7 \end{aligned}$ The $4^{\text{th}}$ term is $7$.